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The given sum is a power series, and it has the form `sum_( n = 1 )^(oo) a_n x^n ` for `a_n = n^2 / ( 2 n ) = n / 2 .`

The question is for what values of `x ` the series converges and what is its sum. The ratio test for a power series shows that this series converges for `-1 lt x lt 1 : ` `| a_( n + 1 ) / a_n | = ( n + 1 ) / n -gt 1 . ` Therefore, the radius of convergence is 1. Note that for x = -1 and x = 1, the series diverges, because its term does not tend to zero.

Now is time to determine the value of this sum. Denote it as `f ( x ) ; ` then

`f ( x ) / x = 1/2 sum_( n = 1 )^(oo) n x^( n - 1 ) .`

It is known that we may integrate power series term-wise inside its interval of convergence, so we obtain

`int ( f ( x ) ) / x dx = 1/2 sum_( n = 1 )^(oo) x^n = 1/2 * x / ( 1 - x ) = 1/2 ( -1 + 1/(1-x) ).`

From this, we get `( f ( x ) ) / x = ( 1 / 2 ( -1 + 1/(1-x) ))' = 1/(2(1-x)^2) `, and finally

`f(x) = x/(2(1-x)^2) .`

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